A proton is moving perpendicular to a uniform magnetic field of 2.5 tesla with 2MeV kinetic energy. The force on the proton is ______ .

Options

A. `8xx10^(-11)`
B. `3xx10^(-10)`
C. `3xx10^(-11)`
D. `8xx10^(-12)`

Answer

A

Explanation

`k=1/2mv^(2)`
`thereforev=sqrt((2k)/m)`
Now, F = Bqv
= `Bexxsqrt((2k)/m)`
= `2.5xx1.6xx10^(-19)sqrt((2xx2xx10^(6)xx1.6xx10^(-19))/(1.6xx10^(-27)))`
= `4xx10^(-19)xxsqrt(4xx10^(6)xx10^(8))`
= `4xx10^(-19)xx2xx10^(7)`
`thereforeF=8xx10^(-12)N`